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Fast way to place bits for puzzle - python

There is a puzzle which I am writing code to solve that goes as follows.
Consider a binary vector of length n that is initially all zeros. You choose a bit of the vector and set it to 1. Now a process starts that sets the bit that is the greatest distance from any 1 bit to $1$ (or an arbitrary choice of furthest bit if there is more than one). This happens repeatedly with the rule that no two 1 bits can be next to each other. It terminates when there is no more space to place a 1 bit. The goal is to place the initial 1 bit so that as many bits as possible are set to 1 on termination.
Say n = 2. Then wherever we set the bit we end up with exactly one bit set.
For n = 3, if we set the first bit we get 101 in the end. But if we set the middle bit we get 010 which is not optimal.
For n = 4, whichever bit we set we end up with two set.
For n = 5, setting the first gives us 10101 with three bits set in the end.
For n = 7, we need to set the third bit to get 1010101 it seems.
I have written code to find the optimal value but it does not scale well to large n. My code starts to get slow around n = 1000 but I would like to solve the problem for n around 1 million.
#!/usr/bin/python
from __future__ import division
from math import *
def findloc(v):
count = 0
maxcount = 0
id = -1
for i in xrange(n):
if (v[i] == 0):
count += 1
if (v[i] == 1):
if (count > maxcount):
maxcount = count
id = i
count = 0
#Deal with vector ending in 0s
if (2*count >= maxcount and count >= v.index(1) and count >1):
return n-1
#Deal with vector starting in 0s
if (2*v.index(1) >= maxcount and v.index(1) > 1):
return 0
if (maxcount <=2):
return -1
return id-int(ceil(maxcount/2))
def addbits(v):
id = findloc(v)
if (id == -1):
return v
v[id] = 1
return addbits(v)
#Set vector length
n=21
max = 0
for i in xrange(n):
v = [0]*n
v[i] = 1
v = addbits(v)
score = sum([1 for j in xrange(n) if v[j] ==1])
# print i, sum([1 for j in xrange(n) if v[j] ==1]), v
if (score > max):
max = score
print max

Latest answer (O(log n) complexity)
If we believe the conjecture by templatetypedef and Aleksi Torhamo (update: proof at the end of this post), there is a closed form solution count(n) calculable in O(log n) (or O(1) if we assume logarithm and bit shifting is O(1)):
Python:
from math import log
def count(n): # The count, using position k conjectured by templatetypedef
k = p(n-1)+1
count_left = k/2
count_right = f(n-k+1)
return count_left + count_right
def f(n): # The f function calculated using Aleksi Torhamo conjecture
return max(p(n-1)/2 + 1, n-p(n-1))
def p(n): # The largest power of 2 not exceeding n
return 1 << int(log(n,2)) if n > 0 else 0
C++:
int log(int n){ // Integer logarithm, by counting the number of leading 0
return 31-__builtin_clz(n);
}
int p(int n){ // The largest power of 2 not exceeding n
if(n==0) return 0;
return 1<<log(n);
}
int f(int n){ // The f function calculated using Aleksi Torhamo conjecture
int val0 = p(n-1);
int val1 = val0/2+1;
int val2 = n-val0;
return val1>val2 ? val1 : val2;
}
int count(int n){ // The count, using position k conjectured by templatetypedef
int k = p(n-1)+1;
int count_left = k/2;
int count_right = f(n-k+1);
return count_left + count_right;
}
This code can calculate the result for n=100,000,000 (and even n=1e24 in Python!) correctly in no time1.
I have tested the codes with various values for n (using my O(n) solution as the standard, see Old Answer section below), and they still seem correct.
This code relies on the two conjectures by templatetypedef and Aleksi Torhamo2. Anyone wants to proof those? =D (Update 2: PROVEN)
1By no time, I meant almost instantly
2The conjecture by Aleksi Torhamo on f function has been empirically proven for n<=100,000,000
Old answer (O(n) complexity)
I can return the count of n=1,000,000 (the result is 475712) in 1.358s (in my iMac) using Python 2.7. Update: It's 0.198s for n=10,000,000 in C++. =)
Here is my idea, which achieves O(n) time complexity.
The Algorithm
Definition of f(n)
Define f(n) as the number of bits that will be set on bitvector of length n, assuming that the first and last bit are set (except for n=2, where only the first or last bit is set). So we know some values of f(n) as follows:
f(1) = 1
f(2) = 1
f(3) = 2
f(4) = 2
f(5) = 3
Note that this is different from the value that we are looking for, since the initial bit might not be at the first or last, as calculated by f(n). For example, we have f(7)=3 instead of 4.
Note that this can be calculated rather efficiently (amortized O(n) to calculate all values of f up to n) using the recurrence relation:
f(2n) = f(n)+f(n+1)-1
f(2n+1) = 2*f(n+1)-1
for n>=5, since the next bit set following the rule will be the middle bit, except for n=1,2,3,4. Then we can split the bitvector into two parts, each independent of each other, and so we can calculate the number of bits set using f( floor(n/2) ) + f( ceil(n/2) ) - 1, as illustrated below:
n=11 n=13
10000100001 1000001000001
<----> <----->
f(6)<----> f(7) <----->
f(6) f(7)
n=12 n=14
100001000001 10000010000001
<----> <----->
f(6)<-----> f(7) <------>
f(7) f(8)
we have the -1 in the formula to exclude the double count of the middle bit.
Now we are ready to count the solution of original problem.
Definition of g(n,i)
Define g(n,i) as the number of bits that will be set on bitvector of length n, following the rules in the problem, where the initial bit is at the i-th bit (1-based). Note that by symmetry the initial bit can be anywhere from the first bit up to the ceil(n/2)-th bit. And for those cases, note that the first bit will be set before any bit in between the first and the initial, and so is the case for the last bit. Therefore the number of bit set in the first partition and the second partition is f(i) and f(n+1-i) respectively.
So the value of g(n,i) can be calculated as:
g(n,i) = f(i) + f(n+1-i) - 1
following the idea when calculating f(n).
Now, to calculate the final result is trivial.
Definition of g(n)
Define g(n) as the count being looked for in the original problem. We can then take the maximum of all possible i, the position of initial bit:
g(n) = maxi=1..ceil(n/2)(f(i) + f(n+1-i) - 1)
Python code:
import time
mem_f = [0,1,1,2,2]
mem_f.extend([-1]*(10**7)) # This will take around 40MB of memory
def f(n):
global mem_f
if mem_f[n]>-1:
return mem_f[n]
if n%2==1:
mem_f[n] = 2*f((n+1)/2)-1
return mem_f[n]
else:
half = n/2
mem_f[n] = f(half)+f(half+1)-1
return mem_f[n]
def g(n):
return max(f(i)+f(n+1-i)-1 for i in range(1,(n+1)/2 + 1))
def main():
while True:
n = input('Enter n (1 <= n <= 10,000,000; 0 to stop): ')
if n==0: break
start_time = time.time()
print 'g(%d) = %d, in %.3fs' % (n, g(n), time.time()-start_time)
if __name__=='__main__':
main()
Complexity Analysis
Now, the interesting thing is, what is the complexity of calculating g(n) with the method described above?
We should first note that we iterate over n/2 values of i, the position of initial bit. And in each iteration we call f(i) and f(n+1-i). Naive analysis will lead to O(n * O(f(n))), but actually we used memoization on f, so it's much faster than that, since each value of f(i) is calculated only once, at most. So the complexity is actually added by the time required to calculate all values of f(n), which would be O(n + f(n)) instead.
So what's the complexity of initializing f(n)?
We can assume that we precompute every value of f(n) first before calculating g(n). Note that due to the recurrence relation and the memoization, generating the whole values of f(n) takes O(n) time. And the next call to f(n) will take O(1) time.
So, the overall complexity is O(n+n) = O(n), as evidenced by this running time in my iMac for n=1,000,000 and n=10,000,000:
> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 1.358s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 13.484s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 3.072s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
And as a by-product of memoization, the calculation of lesser value of n will be much faster after the first call to large n, as you can also see in the sample run. And with language better suited for number crunching such as C++, you might get significantly faster running time
I hope this helps. =)
The code using C++, for performance improvement
The result in C++ is about 68x faster (measured by clock()):
> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 0.020s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 0.198s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 0.047s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
Code in C++:
#include <cstdio>
#include <cstring>
#include <ctime>
int mem_f[10000001];
int f(int n){
if(mem_f[n]>-1)
return mem_f[n];
if(n%2==1){
mem_f[n] = 2*f((n+1)/2)-1;
return mem_f[n];
} else {
int half = n/2;
mem_f[n] = f(half)+f(half+1)-1;
return mem_f[n];
}
}
int g(int n){
int result = 0;
for(int i=1; i<=(n+1)/2; i++){
int cnt = f(i)+f(n+1-i)-1;
result = (cnt > result ? cnt : result);
}
return result;
}
int main(){
memset(mem_f,-1,sizeof(mem_f));
mem_f[0] = 0;
mem_f[1] = mem_f[2] = 1;
mem_f[3] = mem_f[4] = 2;
clock_t start, end;
while(true){
int n;
printf("Enter n (1 <= n <= 10,000,000; 0 to stop): ");
scanf("%d",&n);
if(n==0) break;
start = clock();
int result = g(n);
end = clock();
printf("g(%d) = %d, in %.3fs\n",n,result,((double)(end-start))/CLOCKS_PER_SEC);
}
}
Proof
note that for the sake of keeping this answer (which is already very long) simple, I've skipped some steps in the proof
Conjecture of Aleksi Torhamo on the value of f
For `n>=1`, prove that:
f(2n+k) = 2n-1+1 for k=1,2,…,2n-1 ...(1)
f(2n+k) = k for k=2n-1+1,…,2n ...(2)
given f(0)=f(1)=f(2)=1
The result above can be easily proven using induction on the recurrence relation, by considering the four cases:
Case 1: (1) for even k
Case 2: (1) for odd k
Case 3: (2) for even k
Case 4: (2) for odd k
Suppose we have the four cases proven for n. Now consider n+1.
Case 1:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=1,…,2n-1
= 2n-1+1 + 2n-1+1 - 1
= 2n+1
Case 2:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=0,…,2n-1-1
= 2*(2n-1+1) - 1
= 2n+1
Case 3:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=2n-1+1,…,2n
= i + (i+1) - 1
= 2i
Case 4:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=2n-1+1,…,2n-1
= 2*(i+1) - 1
= 2i+1
So by induction the conjecture is proven.
Conjecture of templatetypedef on the best position
For n>=1 and k=1,…,2n, prove that g(2n+k) = g(2n+k, 2n+1)
That is, prove that placing the first bit on the 2n+1-th position gives maximum number of bits set.
The proof:
First, we have
g(2n+k,2n+1) = f(2n+1) + f(k-1) - 1
Next, by the formula of f, we have the following equalities:
f(2n+1-i) = f(2n+1), for i=-2n-1,…,-1
f(2n+1-i) = f(2n+1)-i, for i=1,…,2n-2-1
f(2n+1-i) = f(2n+1)-2n-2, for i=2n-2,…,2n-1
and also the following inequality:
f(k-1+i) <= f(k-1), for i=-2n-1,…,-1
f(k-1+i) <= f(k-1)+i , for i=1,…,2n-2-1
f(k-1+i) <= f(k-1)+2n-2, for i=2n-2,…,2n-1
and so we have:
f(2n+1-i)+f(k-1+i) <= f(2n+1)+f(k-1), for i=-2n-1,…,2n-1
Now, note that we have:
g(2n+k) = maxi=1..ceil(2n-1+1-k/2)(f(i) + f(2n+k+1-i) - 1)
<= f(2n+1) + f(k-1) - 1
= g(2n+k,2n+1)
And so the conjecture is proven.

So in a break with my normal tradition of not posting algorithms I don't have a proof for, I think I should mention that there's an algorithm that appears to be correct for numbers up to 50,000+ and runs in O(log n) time. This is due to Sophia Westwood, who I worked on this problem with for about three hours today. All credit for this is due to her. Empirically it seems to work beautifully, and it's much, much faster than the O(n) solutions.
One observation about the structure of this problem is that if n is sufficiently large (n ≥ 5), then if you put a 1 anywhere, the problem splits into two subproblems, one to the left of the 1 and one to the right. Although the 1s might be placed in the different halves at different times, the eventual placement is the same as if you solved each half separately and combined them back together.
The next observation is this: suppose you have an array of size 2k + 1 for some k. In that case, suppose that you put a 1 on either side of the array. Then:
The next 1 is placed on the other side of the array.
The next 1 is placed in the middle.
You now have two smaller subproblems of size 2k-1 + 1.
The important part about this is that the resulting bit pattern is an alternating series of 1s and 0s. For example:
For 5 = 4 + 1, we get 10101
For 9 = 8 + 1, we get 101010101
For 17 = 16 + 1, we get 10101010101010101
The reason this matters is the following: suppose you have n total elements in the array and let k be the largest possible value for which 2k + 1 ≤ n. If you place the 1 at position 2k + 1, then the left part of the array up to that position will end up getting tiled with alternating 1s and 0s, which puts a lot of 1s into the array.
What's not obvious is that placing the 1 bit there, for all numbers up to 50,000, appears to yield an optimal solution! I've written a Python script that checks this (using a recurrence relation similar to the one #justhalf) and it seems to work well. The reason that this fact is so useful is that it's really easy to compute this index. In particular, if 2k + 1 ≤ n, then 2k ≤ n - 1, so k ≤ lg (n - 1). Choosing the value ⌊lg (n - 1) ⌋ as your choice of k then lets you compute the bit index by computing 2k + 1. This value of k can be computed in O(log n) time and the exponentiation can be done in O(log n) time as well, so the total runtime is Θ(log n).
The only issue is that I haven't formally proven that this works. All I know is that it's right for the first 50,000 values we've tried. :-)
Hope this helps!

I'll attach what I have. Same as yours, alas, time is basically O(n**3). But at least it avoids recursion (etc), so won't blow up when you get near a million ;-) Note that this returns the best vector found, not the count; e.g.,
>>> solve(23)
[6, 0, 11, 0, 1, 0, 0, 10, 0, 5, 0, 9, 0, 3, 0, 0, 8, 0, 4, 0, 7, 0, 2]
So it also shows the order in which the 1 bits were chosen. The easiest way to get the count is to pass the result to max().
>>> max(solve(23))
11
Or change the function to return maxsofar instead of best.
If you want to run numbers on the order of a million, you'll need something radically different. You can't even afford quadratic time for that (let alone this approach's cubic time). Unlikely to get such a huge O() improvement from fancier data structures - I expect it would require deeper insight into the mathematics of the problem.
def solve(n):
maxsofar, best = 1, [1] + [0] * (n-1)
# by symmetry, no use trying starting points in last half
# (would be a mirror image).
for i in xrange((n + 1)//2):
v = [0] * n
v[i] = count = 1
# d21[i] = distance to closest 1 from index i
d21 = range(i, 0, -1) + range(n-i)
while 1:
d, j = max((d, j) for j, d in enumerate(d21))
if d >= 2:
count += 1
v[j] = count
d21[j] = 0
k = 1
while j-k >= 0 and d21[j-k] > k:
d21[j-k] = k
k += 1
k = 1
while j+k < n and d21[j+k] > k:
d21[j+k] = k
k += 1
else:
if count > maxsofar:
maxsofar = count
best = v[:]
break
return best

Related

I have created a recursive function to calculate the total numbers divisible by n within a range of (start,end). The function works for smaller numbers except when they start getting larger for ex. start=1 and end=10**12 - 1 gives me an error saying that the maximum recursion depth has been exceeded. How do I fix my code to stop this error:
def count(start, end, n, tot=0):
if start > end:
return tot
else:
if start % n == 0:
tot += 1
return count(start + 1, end, n, tot)
start = 1
end = 10**12 - 1
n = 5
print(count(start, end, n))

You can solve this problem with simple arithmetic. Thinking about a range such as (3, 18) (inclusive) and an n of 5, the actual range of interest is from 5 to 15 (since there are no multiples of 5 below 5 or above 15. You can round the start and end values to the new endpoints like this:
start = (start + 4) // n * n
end = end // n * n
What you need to do then is just count the values from 5 to 15 (3) which you can do subtracting the new start from the new end, dividing by n and adding 1 i.e.
(end // n * n - (start + 4) // n * n) // n + 1
which can be simplified to
end // n - (start + 4) // n + 1
So in total your function becomes:
def count(start, end, n):
return end // n - (start + 4) // n + 1

Two points.
(directly answering your question) You can increase the recursion depth by using the sys module.
import sys
sys.setrecursionlimit(10**12)
(towards making you a better programmer) You don't need to use recursion for this problem. In fact, it's a really bad idea as others have noted. Recursion like what you would be looking to do is O(n^2) which is, also known as 'really bad'. A linear approach would be O(n) which is pretty good. Better yet is to take the mathematical approach as Samwise and btilly suggested because you arent tasked to return the numbers that meet the criteria.

I am looking to improve the following simple function (written in python), calculating the maximum size of a specific graph:
def max_size_BDD(n):
i = 1
size = 2
while i <= n:
size += min(pow(2, i-1), pow(2, pow(2, n-i+1))-pow(2, pow(2, n-i)))
i+=1
print(str(i)+" // "+ str(size))
return size
if i give it as input n = 45, the process gets killed (probably because it takes too long, i dont think it is a memory thing, right?). How can i redesign the following algorithm such that it can handle larger inputs?

My proposal: While the original function starts to run into troubles at ~10, I have practically no limitations (even for n = 100000000, I stay below 1s).
def exp_base_2(n):
return 1 << n
def max_size_bdd(n):
# find i at which the min branch switches
start_i = n
while exp_base_2(n - start_i + 1) < start_i:
start_i -= 1
# evaluate all to that point
size = 1 + 2 ** start_i
# evaluate remaining terms (in an uncritical range of n - i)
for i in range(start_i + 1, n + 1):
val = exp_base_2(exp_base_2(n - i))
size += val * (val - 1)
print(f"{i} // {size}")
return size
Remarks:
Core idea: avoid the large powers of 2, as they are not necessary to calculate if you use the min in the end.
I did all this in a rush, maybe I can add more explanation later... if anyone is interested. Then, I could also do a more decent verification of the new implementation.
The effect of exp_base_2 should be negligible after doing all the math to optimize the original calculations. I did this optimization before I went into analysis.
Maybe a complete closed-form solution is possible. I did not invest the time for further investigations.

I need to create function which will take one argument int and output int which represents the number of distinct parts of input integer's partition. Namely,
input:3 -> output: 1 -> {1, 2}
input:6 -> output: 3 -> {1, 2, 3}, {2, 4}, {1, 5}
...
Since I am looking only for distinct parts, something like this is not allowed:
4 -> {1, 1, 1, 1} or {1, 1, 2}
So far I have managed to come up with some algorithms which would find every possible combination, but they are pretty slow and effective only until n=100 or so.
And since I only need number of combinations not the combinations themselves Partition Function Q should solve the problem.
Does anybody know how to implement this efficiently?
More information about the problem: OEIS, Partition Function Q
EDIT:
To avoid any confusion, the DarrylG answer also includes the trivial (single) partition, but this does not affect the quality of it in any way.
EDIT 2:
The jodag (accepted answer) does not include trivial partition.

Tested two algorithms
Simple recurrence relation
WolframMathword algorithm (based upon Georgiadis, Kediaya, Sloane)
Both implemented with Memoization using LRUCache.
Results: WolframeMathword approach orders of magnitude faster.
1. Simple recurrence relation (with Memoization)
Reference
Code
#lru_cache(maxsize=None)
def p(n, d=0):
if n:
return sum(p(n-k, n-2*k+1) for k in range(1, n-d+1))
else:
return 1
Performance
n Time (sec)
10 time elapsed: 0.0020
50 time elapsed: 0.5530
100 time elapsed: 8.7430
200 time elapsed: 168.5830
2. WolframMathword algorithm
(based upon Georgiadis, Kediaya, Sloane)
Reference
Code
# Implementation of q recurrence
# https://mathworld.wolfram.com/PartitionFunctionQ.html
class PartitionQ():
def __init__(self, MAXN):
self.MAXN = MAXN
self.j_seq = self.calc_j_seq(MAXN)
#lru_cache
def q(self, n):
" Q strict partition function "
assert n < self.MAXN
if n == 0:
return 1
sqrt_n = int(sqrt(n)) + 1
temp = sum(((-1)**(k+1))*self.q(n-k*k) for k in range(1, sqrt_n))
return 2*temp + self.s(n)
def s(self, n):
if n in self.j_seq:
return (-1)**self.j_seq[n]
else:
return 0
def calc_j_seq(self, MAX_N):
""" Used to determine if n of form j*(3*j (+/-) 1) / 2
by creating a dictionary of n, j value pairs "
result = {}
j = 0
valn = -1
while valn <= MAX_N:
jj = 3*j*j
valp, valn = (jj - j)//2, (jj+j)//2
result[valp] = j
result[valn] = j
j += 1
return result
Performance
n Time (sec)
10 time elapsed: 0.00087
50 time elapsed: 0.00059
100 time elapsed: 0.00125
200 time elapsed: 0.10933
Conclusion: This algorithm is orders of magnitude faster than the simple recurrence relationship
Algorithm
Reference

I think a straightforward and efficient way to solve this is to explicitly compute the coefficient of the generating function from the Wolfram PartitionsQ link in the original post.
This is a pretty illustrative example of how to construct generating functions and how they can be used to count solutions. To start, we recognize that the problem may be posed as follows:
Let m_1 + m_2 + ... + m_{n-1} = n where m_j = 0 or m_j = j for all j.
Q(n) is the number of solutions of the equation.
We can find Q(n) by constructing the following polynomial (i.e. the generating function)
(1 + x)(1 + x^2)(1 + x^3)...(1 + x^(n-1))
The number of solutions is the number of ways the terms combine to make x^n, i.e. the coefficient of x^n after expanding the polynomial. Therefore, we can solve the problem by simply performing the polynomial multiplication.
def Q(n):
# Represent polynomial as a list of coefficients from x^0 to x^n.
# G_0 = 1
G = [int(g_pow == 0) for g_pow in range(n + 1)]
for k in range(1, n):
# G_k = G_{k-1} * (1 + x^k)
# This is equivalent to adding G shifted to the right by k to G
# Ignore powers greater than n since we don't need them.
G = [G[g_pow] if g_pow - k < 0 else G[g_pow] + G[g_pow - k] for g_pow in range(n + 1)]
return G[n]
Timing (average of 1000 iterations)
import time
print("n Time (sec)")
for n in [10, 50, 100, 200, 300, 500, 1000]:
t0 = time.time()
for i in range(1000):
Q(n)
elapsed = time.time() - t0
print('%-5d%.08f'%(n, elapsed / 1000))
n Time (sec)
10 0.00001000
50 0.00017500
100 0.00062900
200 0.00231200
300 0.00561900
500 0.01681900
1000 0.06701700

You can memoize the recurrences in equations 8, 9, and 10 in the mathematica article you linked for a quadratic in N runtime.

def partQ(n):
result = []
def rec(part, tgt, allowed):
if tgt == 0:
result.append(sorted(part))
elif tgt > 0:
for i in allowed:
rec(part + [i], tgt - i, allowed - set(range(1, i + 1)))
rec([], n, set(range(1, n)))
return result
The work is done by the rec internal function, which takes:
part - a list of parts whose sum is always equal to or less than the target n
tgt - the remaining partial sum that needs to be added to the sum of part to get to n
allowed - a set of number still allowed to be used in the full partitioning
When tgt = 0 is passed, that meant the sum of part if n, and the part is added to the result list. If tgt is still positive, each of the allowed numbers is attempted as an extension of part, in a recursive call.

I found this task and completely stuck with its solution.
A non-empty zero-indexed string S consisting of Q characters is given. The period of this string is the smallest positive integer P such that:
P ≤ Q / 2 and S[K] = S[K+P] for 0 ≤ K < Q − P.
For example, 7 is the period of “abracadabracadabra”. A positive integer M is the binary period of a positive integer N if M is the period of the binary representation of N.
For example, 1651 has the binary representation of "110011100111". Hence, its binary period is 5. On the other hand, 102 does not have a binary period, because its binary representation is “1100110” and it does not have a period.
Consider above scenarios & write a function in Python which will accept an integer N as the parameter. Given a positive integer N, the function returns the binary period of N or −1 if N does not have a binary period.
The attached code is still incorrect on some inputs (9, 11, 13, 17 etc). The goal is to find and fix the bugs in the implementation. You can modify at most 2 line.
def binary_period(n):
d = [0] * 30
l = 0
while n > 0:
d[l] = n % 2
n //= 2
l += 1
for p in range(1, 1 + l):
ok = True
for i in range(l - p):
if d[i] != d[i + p]:
ok = False
break
if ok:
return p
return -1

I was given this piece of code in an interview.
The aim of the exercice is to see where lies the bug.
As an input of the function, you will type the integer to see the binary period of it. As an example solution(4) will give you a binary number of 0011.
However, the question is the following: What is the bug?
The bug in this occasion is not some crash and burn code, rather a behavior that should happen and in the code, do not happen.
It is known as a logical error in the code. Logical error is the error when code do not break but doesn't fullfill the requirements.
Using a brute force on the code will not help as there are a billion possibilities.
However if you run the code, let's say from solutions(1) to solutions(100), you will see that the code runs without any glitch. Yet if you are looking at the code, it should return -1 if there are errors.
The code is not givin any -1 even if you run solutions to a with bigger number like 10000.
The bug here lies in the -1 that is not being triggered.
So let's go step by step on the code.
Could it be the while part?
while n > 0:
d[l] = n % 2
n //= 2
l += 1
If you look at the code, it is doing what it should be doing, changing the number given to a binary number, even if it is doing from a backward position. Instead of having 1011, you have 1101 but it does the job.
The issue lies rather in that part
for p in range(1, 1 + l):
ok = True
for i in range(l - p):
if d[i] != d[i + p]:
ok = False
break
if ok:
return p
return -1
It is not returning -1.
if you put some print on some part of the code like this, this would give you this
for p in range(1, 1 + l):
ok = True
for i in range(l - p):
print('l, which works as an incrementor is substracted to p of the first loop',p,l-p)
if d[i] != d[i + p]:
ok = False
break
if ok:
return p
return -1
If you run the whole script, actually, you can see that it is never ending even if d[i] is not equal anymore to d[i+p].
But why?
The reason is because l, the incrementor was built on an integer division. Because of that, you need to do a 1+l//2.
Which gives you the following
def solution(n):
d = [0] * 30
l = 0
while n > 0:
d[l] = n % 2
n //= 2
l += 1
for p in range(1, 1 + l//2): #here you put l//2
ok = True
print('p est ',p)
for i in range(l - p):
if d[i] != d[i + p]:
ok = False
break
if ok:
return
Now if you run the code with solutions(5) for example, the bug should be fixed and you should have -1.
Addendum:
This test is a difficult one with a not easy algorithm to deal with in very short time, with variables that does not make any sense.
First step would be to ask the following questions:
What is the input of the algorithm? In this case, it is an integer.
What is the expected output? In this case, a -1
Is it a logical error or a crash and burn kind of error? In this case, it is a logical error.
These step-by-step (heuristic) will set you on the right direction to debug a problem.

Following up Andy's solution and checking #hdlopez comment, there is a border case when passing int.MaxVal=2147483647
and if you do not increase the array size to 31 (instead of 30). The function throws an index out of range, so two places need to be modified:
1- int[] d = new int[31]; //changed 30 to 31 (unsigned integer)
2- for (p = 1; p < 1 + l / 2; ++p) //added division to l per statement, P ≤ Q / 2

Here's the challenge I'm failing a hidden test on:
Given integers n, l and r, find the number of ways to represent n as a sum of two integers A and B such that l ≤ A ≤ B ≤ r.
Here's my code:
def countSumOfTwoRepresentations2(n, l, r):
#initializing return variable
totalcount = 0
#returns 0 in impossible cases
if (l + r > n):
return 0
if (r + r < n):
return 0
if (r < (n/2)) or (l > (n/2)):
return 0
# finding the total number of possibilities
p = n / 2
# finding which if l or r is further from 0 or n, respectively
c = max((n-r),l)
# removing impossible cases
totalcount = (p - c) + 1
return totalcount
It passes all the tests I can see the inputs for (and every custom one I could think of), but fails on one of the hidden ones. Any obvious flaws I'm missing? Thanks

As user2357112 noted, you have rejected some valid cases, such as 11, 5, 9: that still permits 6+5 as a solution.
You have also failed to reject an invalid case due to integer division, such as 11, 2, 5, where n == 2*r + 1; however, in these cases, your calculations naturally return 0.
I'm wrong here: r + r < n will catch this case, but this means that your r < (n/2) case is redundant
If you still have trouble with your testing, please include your test classes and test vectors (input sets).